Develop Reference

Check if any part of a string input is not a number

I couldnt find an answer for this in Java, so I'll ask here. I need to check if 3 parts of a string input contains a number (int).
The input will be HOURS:MINUTES:SECONDS (E.g. 10:40:50, which will be 10 hours, 40 minutes and 50 seconds). So far I am getting the values in String[] into an array by splitting it on :. I have parsed the strings into ints and I am using an if statement to check if all 3 parts is equal or larger than 0. The problem is that if I now use letters I will only just get an error, but I want to check if any of the 3 parts contains a character that is not 0-9, but dont know how.
First I thought something like this could work, but really dont.
String[] inputString = input.split(":");
if(inputString.length == 3) {
String[] alphabet = {"a","b","c"};
if(ArrayUtils.contains(alphabet,input)){
gives error message
}
int hoursInt = Integer.parseInt(inputString[0]);
int minutesInt = Integer.parseInt(inputString[1]);
int secondsInt = Integer.parseInt(inputString[2]);
else if(hoursInt >= 0 || minutesInt >= 0 || secondsInt >= 0) {
successfull
}
else {
gives error message
}
else {
gives error message
}
In the end I just want to check if any of the three parts contains a character, and if it doesnt, run something.

If you are sure you always have to parse a String of the form/pattern HH:mm:ss
(describing a time of day),
you can try to parse it to a LocalTime, which will only work if the parts HH, mm and ss are actually valid integers and valid time values.
Do it like this and maybe catch an Exception for a wrong input String:
public static void main(String[] arguments) {
String input = "10:40:50";
String wrongInput = "ab:cd:ef";
LocalTime time = LocalTime.parse(input);
System.out.println(time.format(DateTimeFormatter.ISO_LOCAL_TIME));
try {
LocalTime t = LocalTime.parse(wrongInput);
} catch (DateTimeParseException dtpE) {
System.err.println("Input not parseable...");
dtpE.printStackTrace();
}
}
The output of this minimal example is
10:40:50
Input not parseable...
java.time.format.DateTimeParseException: Text 'ab:cd:ef' could not be parsed at index 0
at java.time.format.DateTimeFormatter.parseResolved0(DateTimeFormatter.java:1949)
at java.time.format.DateTimeFormatter.parse(DateTimeFormatter.java:1851)
at java.time.LocalTime.parse(LocalTime.java:441)
at java.time.LocalTime.parse(LocalTime.java:426)
at de.os.prodefacto.StackoverflowDemo.main(StackoverflowDemo.java:120)

I would personally create my own helper methods for this, instead of using an external library such as Apache (unless you already plan on using the library elsewhere in the project).
Here is an example of what it could look like:
public static void main(String[] arguments) {
String time = "10:50:45";
String [] arr = time.split(":");
if (containsNumbers(arr)) {
System.out.println("Time contained a number!");
}
//You can put an else if you want something to happen when it is not a number
}
private static boolean containsNumbers(String[] arr) {
for (String s : arr) {
if (!isNumeric(s)) {
return false;
}
}
return true;
}
public static boolean isNumeric(String str) {
return str.matches("-?\\d+(.\\d+)?");
}
containsNumbers will take a String array as an input and use an enhanced for loop to iterate through all the String values, using the other helper method isNumeric that checks if the String is a number or not using regex.
This code has the benefit of not being dependent on Exceptions to handle any of the logic.
You can also modify this code to use a String as a parameter instead of an array, and let it handle the split inside of the method instead of outside.
Note that typically there are better ways to work with date and time, but I thought I would answer your literal question.
Example Runs:
String time = "sd:fe:gbdf";
returns false
String time = "as:12:sda";
returns false
String time = "10:50:45";
returns true

You can check the stream of characters.
If the filter does not detect a non-digit, return "Numeric"
Otherwise, return "Not Numeric"
String str = "922029202s9202920290220";
String result = str.chars()
.filter(c -> !Character.isDigit(c))
.findFirst().isEmpty() ? "Numeric"
: "Not Numeric";
System.out.println(result);

If you want to check with nested loop you can see this proposal:
Scanner scanner = new Scanner(System.in);
String [] inputString = scanner.nextLine().split(":");
for (int i = 0; i < inputString.length; i++) {
String current = inputString[i];
for (int k = 0; k < current.length(); k++) {
if (!Character.isDigit(current.charAt(k))) {
System.out.println("Error");
break;
}
}
}

you could use String.matches method :
String notANum= "ok";
String aNum= "7";
if(notANum.matches("^[0-9]+$") sop("no way!");
if(aNum.matches("^[0-9]+$") sop("yes of course!");
The code above would print :
yes of course
The method accepts a regex, the one in the above exemple is for integers.
EDIT
I would use this instead :
if(input.matches("^\d+:\d+:\d+$")) success;
else error
You don't have to split the string.

I tried to make your code better, take a look. You can use Java regex to validate numbers. also defined range for time so no 24:61:61 values is allowed.
public class Regex {
static boolean range(int timeval,int min,int max)
{
boolean status=false;
if(timeval>=min && timeval<max)
{status=true;}
return status;
}
public static void main(String[] args) {
String regex = "[0-9]{1,2}";
String input ="23:59:59";
String msg="please enter valid time ";
String[] inputString = input.split(":");
if(inputString[0].matches(regex) && inputString[1].matches(regex) && inputString[2].matches(regex) )
{
if(Regex.range(Integer.parseInt(inputString[0]), 00, 24) &&Regex.range(Integer.parseInt(inputString[1]), 00, 60) && Regex.range(Integer.parseInt(inputString[2]), 00, 60))
{msg="converted time = " + Integer.parseInt(inputString[0]) + " : " +Integer.parseInt(inputString[1])+ " : " +Integer.parseInt(inputString[2]) ;}
}
System.out.println(msg);
}
}

Separating an unknown amount of hyphens in java?

Good day, guys,
I'm working on a program which requires me to input a name (E.g Patrick-Connor-O'Neill). The name can be composed of as many names as possible, so not necessarily restricted to solely 3 as seen in the example above.But the point of the program is to return the initials back so in this case PCO. I'm writing to ask for a little clarification. I need to separate the names out from the hyphens first, right? Then I need to take the first character of the names and print that out?
Anyway, my question is basically how do I separate the string if I don't know how much is inputted? I get that if it's only like two terms I would do:
final String s = "Before-After";
final String before = s.split("-")[0]; // "Before"
I did attempt to do the code, and all I have so far is:
import java.util.Scanner;
public class main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String input = scan.nextLine();
String[] x = input.split("-");
int u =0;
for(String i : x) {
String y = input.split("-")[u];
u++;
}
}
}
I'm taking a crash course in programming, so easy concepts are hard for me.Thanks for reading!

You don't need to split it a second time. By doing String[] x = input.split("-"); you have an Array of Strings. Now you can iterate over them which you already do with the enhanced for loop. It should look like this
String[] x = input.split("-");
String initials = "";
for (String name : x) {
initials += name.charAt(0);
}
System.out.println(initials);
Here are some Java Docs for the used methods
String#split
String#charAt
Assignment operator +=

You can do it without splitting the string by using String.indexOf to find the next -; then just append the subsequent character to the initials:
String initials = "" + input.charAt(0);
int next = -1;
while (true) {
next = input.indexOf('-', next + 1);
if (next < 0) break;
initials += input.charAt(next + 1);
}
(There are lots of edge cases not handled here; omitted to get across the main point of the approach).

In your for-each loop append first character of all the elements of String array into an output String to get the initials:
String output = "";
for(String i : x) {
output = output + y.charAt(0);
}

This will help.
public static void main(String[] args) {
String output = "";
String input = "Patrick-Connor-O'Neil-Saint-Patricks-Day";
String[] brokenInput = input.split("-");
for (String temp : brokenInput) {
if (!temp.equals(""))
output = output + temp.charAt(0);
}
System.out.println(output);
}

You could totally try something like this (a little refactor of your code):
import java.util.Scanner;
public class main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String input = "";
System.out.println("What's your name?");
input = scan.nextLine();
String[] x = input.split("-");
int u =0;
for(String i : x) {
String y = input.split("-")[u];
u++;
System.out.println(y);
}
}
}
I think it's pretty easy and straightforward from here if you want to simply isolate the initials. If you are new to Java make sure you use a lot of System.out since it helps you a lot with debugging.
Good coding.
EDIT: You can use #Mohit Tyagi 's answer with mine to achieve the full thing if you are cheating :P

This might help
String test = "abs-bcd-cde-fgh-lik";
String[] splitArray = test.split("-");
StringBuffer stringBuffer = new StringBuffer();
for (int i = 0; i < splitArray.length; i++) {
stringBuffer.append(splitArray[i].charAt(0));
}
System.out.println(stringBuffer);
}
Using StringBuffer will save your memory as, if you use String a new object will get created every time you modify it.

return the first word in any string (Java)

I have to be able to input any two words as a string. Invoke a method that takes that string and returns the first word. Lastly display that word.
The method has to be a for loop method. I kind of know how to use substring, and I know how to return the first word by just using .substring(0,x) x being how long the first word is.
How can I make it so that no matter what phrase I use for the string, it will always return the first word? And please explain what you do, because this is my first year in a CS class. Thank you!

I have to be able to input any two words as a string
The zero, one, infinity design rule says there is no such thing as two. Lets design it to work with any number of words.
String words = "One two many lots"; // This will be our input
and then invoke and display the first word returned from the method,
So we need a method that takes a String and returns a String.
// Method that returns the first word
public static String firstWord(String input) {
return input.split(" ")[0]; // Create array of words and return the 0th word
}
static lets us call it from main without needing to create instances of anything. public lets us call it from another class if we want.
.split(" ") creates an array of Strings delimited at every space.
[0] indexes into that array and gives the first word since arrays in java are zero indexed (they start counting at 0).
and the method has to be a for loop method
Ah crap, then we have to do it the hard way.
// Method that returns the first word
public static String firstWord(String input) {
String result = ""; // Return empty string if no space found
for(int i = 0; i < input.length(); i++)
{
if(input.charAt(i) == ' ')
{
result = input.substring(0, i);
break; // because we're done
}
}
return result;
}
I kind of know how to use substring, and I know how to return the first word by just using .substring(0,x) x being how long the first word is.
There it is, using those methods you mentioned and the for loop. What more could you want?
But how can I make it so that no matter what phrase I use for the string, it will always return the first word?
Man you're picky :) OK fine:
// Method that returns the first word
public static String firstWord(String input) {
String result = input; // if no space found later, input is the first word
for(int i = 0; i < input.length(); i++)
{
if(input.charAt(i) == ' ')
{
result = input.substring(0, i);
break;
}
}
return result;
}
Put it all together it looks like this:
public class FirstWord {
public static void main(String[] args) throws Exception
{
String words = "One two many lots"; // This will be our input
System.out.println(firstWord(words));
}
// Method that returns the first word
public static String firstWord(String input) {
for(int i = 0; i < input.length(); i++)
{
if(input.charAt(i) == ' ')
{
return input.substring(0, i);
}
}
return input;
}
}
And it prints this:
One
Hey wait, you changed the firstWord method there.
Yeah I did. This style avoids the need for a result string. Multiple returns are frowned on by old programmers that never got used to garbage collected languages or using finally. They want one place to clean up their resources but this is java so we don't care. Which style you should use depends on your instructor.
And please explain what you do, because this is my first year in a CS class. Thank you!
What do I do? I post awesome! :)
Hope it helps.

String line = "Hello my name is...";
int spaceIndex = line.indexOf(" ");
String firstWord = line.subString(0, spaceIndex);
So, you can think of line as an array of chars. Therefore, line.indexOf(" ") gets the index of the space in the line variable. Then, the substring part uses that information to get all of the characters leading up to spaceIndex. So, if space index is 5, it will the substring method will return the indexes of 0,1,2,3,4. This is therefore going to return your first word.

The first word is probably the substring that comes before the first space. So write:
int x = input.indexOf(" ");
But what if there is no space? x will be equal to -1, so you'll need to adjust it to the very end of the input:
if (x==-1) { x = input.length(); }
Then use that in your substring method, just as you were planning. Now you just have to handle the case where input is the blank string "", since there is no first word in that case.

Since you did not specify the order and what you consider as a word, I'll assume that you want to check in given sentence, until the first space.
Simply do
int indexOfSpace = sentence.indexOf(" ");
firstWord = indexOfSpace == -1 ? sentence : sentence.substring(0, indexOfSpace);
Note that this will give an IndexOutOfBoundException if there is no space in the sentence.
An alternative would be
String sentences[] = sentence.split(" ");
String firstWord = sentence[0];
Of if you really need a loop,
String firstWord = sentence;
for(int i = 0; i < sentence.length(); i++)
{
if(sentence.charAt(i) == ' ')
{
sentence = firstWord.substring(0, i);
break;
}
}

You may get the position of the 'space' character in the input string using String.indexOf(String str) which returns the index of the first occurrence of the string in passed to the method.
E.g.:
int spaceIndex = input.indexOf(" ");
String firstWord = input.substring(0, spaceIndex);

Maybe this can help you figure out the solution to your problem. Most users on this site don't like doing homework for students, before you ask a question, make sure to go over your ISC book examples. They're really helpful.
String Str = new String("Welcome to Stackoverflow");
System.out.print("Return Value :" );
System.out.println(Str.substring(5) );
System.out.print("Return Value :" );
System.out.println(Str.substring(5, 10) );

Split a string in Java and picking up a part of it

Say I got a string from a text file like
"Yes ABC 123
Yes DEF 456
Yes GHI 789"
I use this code to split the string by whitespace.
while (inputFile.hasNext())
{
String stuff = inputFile.nextLine();
String[] tokens = stuff.split(" ");
for (String s : tokens)
System.out.println(s);
}
But I also want to assign Yes to a boolean, ABC to another string, 123 to a int.
How can I pick them up separately? Thank you!

boolean b=tokens[0].equalsIgnoreCase("yes");
String name=tokens[1];
int i=Integer.parseInt(tokens[2]);

Could you clarify what the exact purpose of what you're doing is? You can refer to the separate Strings with tokens[i] with i being the index. You could throw these into a switch statement (since Java 7) and match for the words you're looking for. Then you can take further action, i.e. convert the Strings to Booleans or Ints.
You should consider checking the input to be valid too even if you are expecting the file to always have those 3 words separated by a space.

Create Class Line and List<Line> that will store all your file into list:
public class Line{
private boolean mFlag = false;
private int mNum = 0;
private String mStr;
public Line(String stuff) {
String[] tokens = stuff.split("[ ]+");
if(tokens.length ==3){
mFlag=tokens[0].equalsIgnoreCase("yes");
mNum=Integer.parseInt(tokens[1]);
mStr=tokens[3];
}
}
}
and call it:
public static void main(String[] args) {
List<Line> list = new ArrayList<Line>();
Line line;
while (inputFile.hasNext())
{
String stuff = inputFile.nextLine();
line = new Line(stuff);
list.add(line);
}
}

If your input String is going to be in the same format always i.e. boolean,String ,int then you can access the individual indices of token array and convert them to your specified format
boolean opinion = tokens[0].equalsIgnoreCase("yes");
String temp = token[1];
int i = Integer.parseInt(token[2])
But you might require to create an array or something that stores the values for consecutive inputs that user does otherwise these variables would be over ridden for every new input from user.

How can i use substring or other function to divide a string into words?

So I have this function:
public void actionPerformed(ActionEvent e)
{
String input = jt.toString();
}
And I want to use substring or any other function to divide the words in this sentence.
Like using substring from 0 until it finds a "space" and then it should stop and then start again until the end of the sentence.

What you need is to use the Split method for the String class.
String[] input = jt.toString().split("\\s+");
This method will give you an array where each cell of the array will contain a word.
jt stands for JText?
If yes, you should get the text from the component instead of converting the object to string using the following instruction: jt.getText()

string[] words = input.Split(' ')

You can also do it with stringbuilder if you care about saving memory.
public class splitword {
public static void main(String[] args) {
String input = "Hi! Can you please split me into pieces :0";
String[] toSplit = new StringBuilder(input).toString().split("[\\s\\p{P}&& [^']]+");
for (String x : toSplit) {
System.out.println(x);
}
}
}